Point-slope and two-point forms

The equa­tion of a line is an equa­tion in $x$ and $y$ that is sat­is­fied by the coor­di­nates of every point on the line and by no other point. With it, ques­tions about the line become alge­bra: to test whether a point lies on the line, sub­sti­tute; to find where two lines meet, solve. This les­son finds the equa­tions of lines par­al­lel to the axes, and of a line given by one point and its slope, or by two of its points.

Lines par­al­lel to the axes#

Every point on a hor­i­zon­tal line has the same $y$-coor­di­nate. So the line par­al­lel to the $x$-axis at a dis­tance $b$ from it is

$$y = b$$

(above the axis if $b > 0$, below if $b < 0$). In par­tic­u­lar, the $x$-axis itself is $y = 0$.

Sim­i­larly, every point on a ver­ti­cal line has the same $x$-coor­di­nate, so the line par­al­lel to the $y$-axis is

$$x = a$$

and the $y$-axis itself is $x = 0$.

For exam­ple, the hor­i­zon­tal line through $(3, -5)$ is $y = -5$, and the ver­ti­cal line through it is $x = 3$.

Point-slope form#

Result. The equa­tion of the line through the point $(x_1, y_1)$ with slope $m$ is

$$y - y_1 = m(x - x_1)$$

Why. Let $P(x, y)$ be any point on the line other than $(x_1, y_1)$. The slope of the line, worked out from these two points, must be $m$:

$$\displaystyle \frac{y - y_1}{x - x_1} = m \implies y - y_1 = m(x - x_1)$$

The point $(x_1, y_1)$ itself also sat­is­fies this equa­tion. Con­versely, any point sat­is­fy­ing it gives slope $m$ with $(x_1, y_1)$, so it lies on the line.

It is usual to give the final answer in the form $ax + by + c = 0$ with whole-num­ber coef­fi­cients.

Exam­ple 1#

Find the equa­tion of the line through $(2, -3)$ with slope $4$.

$$\begin{aligned} y - (-3) &= 4(x - 2) \\ y + 3 &= 4x - 8 \\ 4x - y - 11 &= 0 \end{aligned}$$

Check: at $(2, -3)$, $8 + 3 - 11 = 0$.

Exam­ple 2#

Find the equa­tion of the line through $(3, 4)$ with incli­na­tion $45°$.

The slope is $\tan 45° = 1$, so $y - 4 = 1(x - 3)$, that is, $x - y + 1 = 0$.

Two-point form#

Result. The equa­tion of the line through $(x_1, y_1)$ and $(x_2, y_2)$, where $x_1 \neq x_2$, is

$$\displaystyle y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$$

This is sim­ply the point-slope form with the slope worked out from the two points. (If $x_1 = x_2$, the line is ver­ti­cal, with equa­tion $x = x_1$.)

Exam­ple 3#

Find the equa­tion of the line through $(1, 2)$ and $(5, -6)$.

The slope is $\displaystyle \frac{-6 - 2}{5 - 1} = \frac{-8}{4} = -2$. Then

$$\begin{aligned} y - 2 &= -2(x - 1) \\ y - 2 &= -2x + 2 \\ 2x + y - 4 &= 0 \end{aligned}$$

Check with the sec­ond point: $2(5) + (-6) - 4 = 0$.

Lines in a tri­an­gle#

Two lines con­nected with a tri­an­gle are often asked for.

• A median joins a ver­tex to the mid­point of the oppo­site side.
• The per­pen­dic­u­lar bisec­tor of a seg­ment passes through its mid­point at right angles to it.

Exam­ple 4#

The ver­tices of a tri­an­gle are $A(1, 4)$, $B(3, -2)$ and $C(5, 6)$. Find the equa­tion of the median through $A$.

The mid­point of $BC$ is $\displaystyle \left(\frac{3 + 5}{2}, \frac{-2 + 6}{2}\right) = (4, 2)$. The median joins $A(1, 4)$ to $(4, 2)$, with slope $\displaystyle \frac{2 - 4}{4 - 1} = -\frac{2}{3}$:

$$\displaystyle \begin{aligned} y - 4 &= -\frac{2}{3}(x - 1) \\ 3y - 12 &= -2x + 2 \\ 2x + 3y - 14 &= 0 \end{aligned}$$

Exam­ple 5#

Find the per­pen­dic­u­lar bisec­tor of the seg­ment join­ing $A(1, 3)$ and $B(5, 7)$.

The mid­point is $(3, 5)$. The slope of $AB$ is $\displaystyle \frac{7 - 3}{5 - 1} = 1$, so the per­pen­dic­u­lar slope is $-1$:

$$y - 5 = -(x - 3) \implies x + y - 8 = 0$$

Every point on this line is the same dis­tance from $A$ as from $B$.

Prac­tice#

1. Write the equa­tion of the line par­al­lel to the $x$-axis through $(3, -5)$.
2. Write the equa­tion of the line par­al­lel to the $y$-axis through $(-2, 7)$.
3. Find the equa­tion of the line through $(1, 3)$ with slope $2$.
4. Find the equa­tion of the line through $(-2, 4)$ with slope $\displaystyle -\frac{3}{2}$.
5. Find the equa­tion of the line through the ori­gin with incli­na­tion $150°$.
6. Find the equa­tion of the line through $(2, 5)$ and $(4, 9)$.
7. Find the equa­tion of the line through $(-1, 3)$ and $(3, -5)$.
8. Find the equa­tion of the line through $(3, 1)$ and $(3, 6)$.
9. The ver­tices of a tri­an­gle are $A(2, 5)$, $B(-4, 1)$, $C(6, -3)$. Find the median through $A$.
10. Find the per­pen­dic­u­lar bisec­tor of the seg­ment join­ing $(-1, 2)$ and $(3, 6)$.

Answers#